Problem 6 - Algebra and Trigonometry

Question

\(5-16=\) Solve the given equation. $$ \sin \theta=-\frac{\sqrt{2}}{2} $$

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{5\pi}{4} + 2k\pi \) and \( \theta = \frac{7\pi}{4} + 2k\pi \) for integer \( k \).

Step 1: Identify the reference angle

The given equation is \( \sin \theta = -\frac{\sqrt{2}}{2} \). We first find the reference angle \( \alpha \) such that \( \sin \alpha = \frac{\sqrt{2}}{2} \). This value corresponds to \( \alpha = \frac{\pi}{4} \) or \( 45^\circ \).

Step 2: Determine the quadrants

Since \( \sin \theta = -\frac{\sqrt{2}}{2} \) and sine is negative, \( \theta \) must lie in either the third or fourth quadrant.

Step 3: Find the solutions in the third quadrant

In the third quadrant, the angle \( \theta \) is \( \pi + \alpha \). Thus, \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \), or \( 225^\circ \).

Step 4: Find the solutions in the fourth quadrant

In the fourth quadrant, the angle \( \theta \) is \( 2\pi - \alpha \). Thus, \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \), or \( 315^\circ \).

Step 5: Compile the general solutions

Since sine is a periodic function with period \( 2\pi \), the general solutions are \( \theta = \frac{5\pi}{4} + 2k\pi \) and \( \theta = \frac{7\pi}{4} + 2k\pi \), where \( k \) is any integer.

Key Concepts

Reference Angle

The concept of a reference angle is crucial when solving trigonometric equations. A reference angle, often denoted as \( \alpha \), is the angle formed between the last ray and the horizontal axis which helps identify equivalent angles in different quadrants. In the case of \( \sin \theta = -\frac{\sqrt{2}}{2} \), the aim is to first find the corresponding reference angle for the positive value, \( \sin \alpha = \frac{\sqrt{2}}{2} \). This is a common trigonometric value, and we know that it corresponds to \( \alpha = \frac{\pi}{4} \) or \( 45^\circ \).

This means that any angle \( \theta \) that has a reference angle of \( \frac{\pi}{4} \) will give the same sine value, either positive or negative depending on its quadrant. Using reference angles can significantly simplify the process of finding solutions to trigonometric equations.

Quadrant Analysis

Understanding which quadrant an angle lies in is key when dealing with trigonometric functions. The sign of a trigonometric function depends on the quadrant in which the corresponding angle falls. Sine is negative in the third and fourth quadrants, which is why the solutions to \( \sin \theta = -\frac{\sqrt{2}}{2} \) must exist in these quadrants.

• In the third quadrant, the angle is conventionalized by adding the reference angle to \( \pi \). Therefore, \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \).
• In the fourth quadrant, the angle is obtained by subtracting the reference angle from \( 2\pi \). Consequently, \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).

Remembering the behavior of each trigonometric ratio in the four quadrants makes solving equations more intuitive.

General Solutions

Trigonometric functions are periodic, which means they repeat their values in regular intervals. When we talk about the general solution of a trigonometric equation, we refer to all possible angles that satisfy the given equation, taking into account the period of the function.

For sine functions, this period is \( 2\pi \). Therefore, any angle solution can have multiples of \( 2\pi \) added or subtracted. If \( \theta \) is a solution, then other solutions could be \( \theta + 2k\pi \) where \( k \) is any integer.

Thus, for our equations:

• In the third quadrant the general solution is \( \theta = \frac{5\pi}{4} + 2k\pi \).
• In the fourth quadrant, the general solution is \( \theta = \frac{7\pi}{4} + 2k\pi \).

This ensures that every possible angle that meets the condition \( \sin \theta = -\frac{\sqrt{2}}{2} \) is captured. Understanding the concept of general solutions helps in recognizing the infinity of solutions beyond just the principal value.